The number of ways in which a team of 11 players can be formed out of 25 players, if 6 out of them are always to be included and 5 of them are always to be excluded, is

Given 25 players, with 6 players always included and 5 players always excluded, to form an 11-player team, we first exclude the 5 players who cannot participate and note that 6 players are fixed. The remaining 5 positions must be chosen from the effective pool of $25 - 6 - 5 = 14$ players.

The selection is a combination: $\binom{14}{5} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002$.

Final answer: $2002$.